Does L'Hospital's rule apply to complex functions?

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Discussion Overview

The discussion centers around the application of L'Hospital's rule to limits involving complex functions, specifically evaluating the limit of a ratio as \( x \) approaches infinity. The context includes elements of quantum mechanics and complex analysis.

Discussion Character

  • Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes evaluating the limit \(\lim_{x\rightarrow \infty} \frac{x-1}{e^{ipx/\hbar}}\) and questions if it equals \(\lim_{x\rightarrow \infty} \frac{\hbar}{ip e^{ipx/\hbar}}\).
  • Another participant argues that L'Hospital's rule does not apply since the limit does not approach \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), noting that \(\lim_{x\rightarrow\infty}e^{ix}\) does not exist due to oscillation.
  • A later reply suggests that the limit does not exist because both the real and imaginary parts oscillate without settling as \( x \) increases.
  • One participant humorously reflects on the complexity of the quantum mechanics problem, indicating a struggle with the application.
  • Another participant asserts that the limit approaches infinity due to the numerator increasing without bound while the denominator remains bounded.
  • One participant recommends choosing different boundary conditions for the quantum mechanics problem to potentially resolve the issue.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of L'Hospital's rule and the behavior of the limit, indicating no consensus on the resolution of the limit's existence or value.

Contextual Notes

There are unresolved assumptions regarding the behavior of the complex exponential function and the implications of boundary conditions in quantum mechanics.

quasar987
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I have to evaluate

[tex]\lim_{x\rightarrow \infty} \frac{x-1}{e^{ipx/\hbar}}[/tex]

is this equal to

[tex]\lim_{x\rightarrow \infty} \frac{\hbar}{ip e^{ipx/\hbar}}[/tex]

??
 
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L'Hospital's rule only applies when a limit approaches [itex]\frac{0}{0}[/itex] or [itex]\frac{\infty}{\infty}[/itex]. Niether is the case here because [itex]\lim_{x\rightarrow\infty}e^{ix}[/itex] does not exist. The real and imaginary parts oscilate between -1 and 1 as x approaches infinity. Your limit is equal to:
[tex]\lim_{x\rightarrow \infty} \frac{x-1}{e^{ipx/\hbar}} =\lim_{x\rightarrow \infty} (x-1)\cos{\frac{ipx}{\hbar}}-i(x-1)\sin{\frac{ipx}{\hbar}[/tex]
So the real and imaginary parts both oscillate between larger and larger positive and negative numbers as x gets larger, so the limit does not exist.
 
Dang. I guess I was trying a little too hard to make this QM problem work :-p
 
ok so the answer is infinity
we have some thing bounded in the denominator
and the numerator goes to infinity
 
I suggest you choose different boundary conditions at infinity for your QM problem.
 

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